Leetcode 初探

Add Two Numbers

第一次在leetcode上面做题，选了两道最简单的题目来试试手，回忆一下cpp的写法。写了一个暑假的python之后，在cpp语法上稍微有点不太习惯，还好，做完前两道题就上手了。

这里主要讲一讲Question2.
题目原意：

You are given two non-empty linked lists representing two non-negative integers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list.
You may assume the two numbers do not contain any leading zero, except the number 0 itself.
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8

这种题目以前做的也挺多，就是两个链表相加，注意进位问题。

第一次写出来的代码是这样的（确实有点久没写c++,有点生疏…)：

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     ListNode *next; *     ListNode(int x) : val(x), next(NULL) {} * }; */class Solution {public:    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {        ListNode* ans = new ListNode(l1->val + l2->val);        ListNode* pre = ans;        ListNode* cur1 = l1->next, *cur2 = l2->next;        int carry = 0;        Carry(ans, carry);        while (cur1 != NULL || cur2 != NULL) {            ListNode* node = new ListNode(0);            if (cur1 != NULL && cur2 != NULL) {                node->val = cur1->val + cur2->val + carry;                cur1 = cur1->next;                cur2 = cur2->next;            }            else if (cur1 != NULL) {                node->val = cur1->val + carry;                cur1 = cur1->next;            }            else if (cur2 != NULL) {                node->val = cur2->val + carry;                cur2 = cur2->next;            }            pre->next = node;            pre = node;            Carry(node, carry);        }        if (carry == 1)             pre->next = new ListNode(1);        return ans;    }    void Carry(ListNode*& node, int &carry) {        if (node->val > 9) {            carry = 1;            node->val -= 10;        }        else             carry = 0;    }};

看了答案后，发现答案的写法确实简洁：

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     ListNode *next; *     ListNode(int x) : val(x), next(NULL) {} * }; */class Solution {public:    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {        ListNode* dummyHead = new ListNode(0);        ListNode* p = l1, * q = l2, *cur = dummyHead;        int carry = 0;        while (p != NULL || q != NULL) {            //利用三元表达式处理if语句            int x = (p != NULL)? p->val: 0;            int y = (q != NULL)? q->val: 0;            int sum = x + y + carry;            //利用除法和取余处理多种情况            carry = sum / 10;            cur->next = new ListNode(sum % 10);            cur = cur->next;            if (p != NULL) p = p->next;            if (q != NULL) q = q->next;        }        //最后处理进位        if (carry)            cur->next = new ListNode(1);        return dummyHead->next;    }};